Vaidev and Sean started on a 48-km cycling trip at the same time. They cycled at the same speed for first 10 km. For the remaining 38 km, Vaidev cycled faster than Sean. He arrived at the finishing point 40 minutes before Sean who was 10 km behind him. Sean did not change his speed throughout and completed it at 09 25.

1. At what time did the trip begin? Give the answer in 12-hour format.
2. What was Vaidev's average speed for the remaining 38 km of the trip in km/h?

(a)
60 min = 1 h
40 min =⁴⁰₆₀ h = ²₃ 

Sean's average speed
= 10 ÷ ²₃
= 15 km/h

Time that Sean took for the trip
= 48 ÷ 15
= 3³₁₅ h
= 3¹₅ h
= 3 h 12 min

3 h 12 min before 09 25 is 06 13.

(b)
To find the time that Vaidev took for the remaining 38 km of the trip, we need to use the time that Sean took subtract that of the time taken for first 10 km and the time that Vaidev was faster than Sean.

Time that Vaidev took for the remaining 38 km of the trip
= 3¹₅ - ²₃ - ²₃
= 1¹³₁₅ h

Vaidev's average speed
= 38 ÷ 1¹³₁₅
= 20⁵₁₄ km/h

Answer(s): (a) 06 13; (b) 20⁵₁₄ km/h